Problem: $\int^{2}_{0}\dfrac{x}{(1+2x^2)^2}\,dx\, = $
Strategy Let's first find the indefinite integral $\int\dfrac{x}{(1+2x^2)^2}\,dx\, $. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int\dfrac{x}{(1+2x^2)^2}\,dx\, $, we can use U-substitution. If we let $ {u=1+2x^2}$, then ${du=4x \, dx}$ and ${ x\,dx=\dfrac{1}{4}\, du}$. So we have: $\begin{aligned}\int\dfrac{x}{(1+2x^2)^2}\,dx\, &=\int\dfrac{1}{({1+2x^2})^2}\,{x\, dx}\,\\\\\\\\ &=\int\dfrac{1}{ u^2}\,\cdot {\dfrac14\, du}\,\\\\\\\\ &=\dfrac14\int\dfrac{1}{u^2}\,du\\\\\\\\ &=\dfrac{1}{4}\cdot -u^{-1}+C\\\\\\\\ &=-\dfrac14\left(1+2x^2\right)^{-1}+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{2}_{0}\dfrac{x}{(1+2x^2)^2}\,dx\, &= -\dfrac14\left(1+2x^2\right)^{-1}\Bigg|^2_0\\\\\\\\ &=-\dfrac{1}{4}\left(\dfrac19-1\right)\\\\\\\\ &=\dfrac29\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{2}_{0}\dfrac{x}{(1+2x^2)^2}\,dx\, = \dfrac29$